find the enthalpy for the combustion of methanol.

Question

The combustion of methanol is shown by the following equation:

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)

Using the equations below, 

1. C(s) + O₂(g) → CO₂(g) ΔH + 393 kJ/mol
2. H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH + 242 kJ/mol
3. C(s) + 2 H₂(g) + 1/2 O₂(g) → CH₃OH(l) ΔH + 638 kJ/mol

(a) find the enthalpy for the combustion of methanol.

  

(b) What mass of water could be heated from 20.0 °C to 35.0 °C by the burning of 2.57 mol of methanol? 

Answer 1

→ I think you have the wrong signs for the enthalpies of formation; they should be negative.

This uses Hess’ law that states the separate equations can be combined in such a way that similar species on each side of the summed equations can cancel.

The equations can be written

2CH3OH → 2C + 4H2  + O2  ΔH →  2 x +638 kj/mol = 1276 kj

2C + 2O2 → 2CO2                  ΔH = 2 x -393 kj/mol = -786 kj

4H2 + 2O2 → 4H2O              ΔH = 4 x -242 kj/mol = -968 kj



Adding these equations and cancelling equal species on each side gives

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)   ΔH = -478 kj or -239 kj/mol

The amount of heat obtained by burning 2.57 moles of methanol is 239 kj/mol x 2.57 moles = 614 kj or 614 x 10^3 j.

As the specific heat of liquid water is 4.186 J/g-deg, the mass of liquid water that can be heated from 20.0 to 35.0 °C is given by

ΔH = 4.186 j/g-deg x m x ( 35 – 20)° = 614 x 10^3  j (m is mass). Solve for mass.

 m = 614 x 10^3 j / (4.186 j/g-°C x 15°C) = 9.78 x 10^3 grams.